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## Prime polynomial

Posted by Chris on November 19, 2011 – 2:50 pm

How many positive integers n < 1000 makes the following be prime?:

3^n + 5^n + 7^n + 11^n + 13^n + 17^n + 19^n

This post is under “MathsChallenge” and has 6 respond so far.

### 6 Responds so far- Add one»

1. 1. slavy Said：

Too easy, Chris

2. 2. Chris Said：

Hi Slavy. True, but it looks hard.

3. 3. Chris Said：

It looks so hard that nobody else is trying. In fact it’s a slightly trick question. The answer is 0 for any integer n > 0. Can anybody show why that’s the case?

4. 4. jan Said：

mod(3^n,3)=0

n is odd
mod(5^n,3)=0
mod(7^n,3)=0
mod(11^n,3)=0
mod(13^n,3)=0
mod(17^n,3)=0
mod(19^n,3)=0

n is even
mod(5^n,3)=1
mod(7^n,3)=1
mod(11^n,3)=1
mod(13^n,3)=1
mod(17^n,3)=1
mod(19^n,3)=1
which adds to 6 and mod(6,3)=0

so the number described is divisible by 3.

5. 5. jan Said：

mod(5^n,3)=-1
mod(7^n,3)=1
mod(11^n,3)=-1
mod(13^n,3)=1
mod(17^n,3)=-1
mod(19^n,3)=1

obv

6. 6. Chris Said：

Hi jan. That’s right. Thanks. I thought that I was going to have to answer it.

For any n, 3^n + 5^n + 7^n + 11^n + 13^n + 17^n + 19^n
= 0^n + (-1)^n + 1^n + (-1)^n + 1^n + (-1)^n + 1^n (mod 3)
= 3*1^n + 3*(-1)^n = 0 (mod 3)
So the “polynomial” is always divisible by 3 and can never be prime.

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